Question 288: Dual Nature of Matter and Radiation

Question

The maximum velocity of the photoelectrons emitted by a metal surface is $9 	imes 10^5 \mathrm{~m} / \mathrm{s}$. The value of ratio of charge (e) to mass (m) of the photoelectron is $1.8 	imes 10^{11} \mathrm{C} / \mathrm{kg}$. The value of stopping potential in volt is

Accepted Answer

Given:
Maximum velocity of photoelectrons, $v_{max} = 9 	imes 10^5~	ext{m/s}$
$\dfrac{e}{m} = 1.8 	imes 10^{11}~	ext{C/kg}$
Let the stopping potential be $V_0$.
Step 1: Kinetic Energy of Photoelectron
The maximum kinetic energy of the photoelectron is given by:
$$ K_{max} = \frac{1}{2}mv_{max}^2 $$
Step 2: Stopping Potential Relation
The stopping potential $V_0$ is related by the equation:
$$ eV_0 = K_{max} $$
So,
$$ V_0 = \frac{K_{max}}{e} $$
Step 3: Substitute for $K_{max}/e$
Substitute $K_{max}$:
$$ V_0 = \frac{\frac{1}{2}mv_{max}^2}{e} $$
But $\frac{e}{m}$ is given.
So,
$$ \frac{m}{e} = \frac{1}{\frac{e}{m}} = \frac{1}{1.8	imes 10^{11}} $$
Now,
$$ V_0 = \frac{1}{2} \cdot \frac{m}{e} (v_{max})^2 $$
Step 4: Substitute Values
Substitute:
$v_{max} = 9 	imes 10^5~	ext{m/s}$
$m/e = \dfrac{1}{1.8 	imes 10^{11}}$
So,
$ V_0 = \frac{1}{2} \left( \frac{1}{1.8 	imes 10^{11}} ight) \left(9 	imes 10^5ight)^2 $
Step 5: Calculate $(9 	imes 10^5)^2$
$ (9 	imes 10^5)^2 = 81 	imes 10^{10} = 8.1 	imes 10^{11} $
Step 6: Substitute and Calculate
$ V_0 = \frac{1}{2} 	imes \frac{1}{1.8 	imes 10^{11}} 	imes (8.1 	imes 10^{11}) $
$ V_0 = \frac{1}{2} 	imes \frac{8.1 	imes 10^{11}}{1.8 	imes 10^{11}} $
$ V_0 = \frac{1}{2} 	imes \frac{8.1}{1.8} $
$ V_0 = \frac{1}{2} 	imes 4.5 = 2.25~	ext{V} $
Final Answer:
Option B: $2.25$
volts

Answer

2.00

Answer

2.25

Answer

2.50

Answer

3.00

Dual Nature of Matter and Radiation

The maximum velocity of the photoelectrons emitted by a metal surface is 9 \times 10^5 \mathrm{~m} / \mathrm{s}. The value of ratio of charge (e) to mass (m) of the photoelectron is 1.8 \times 10^{11} \mathrm{C} / \mathrm{kg}. The value of stopping potential in volt is

Choose the correct answer: